If λ₁ and λ₂ are the wavelengths of the first members of the Lyman… | MHT CET 2026 (14 May, Shift 2)

Q6MHT CET 2026 · 14 MAY · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the first members of the Lyman and Paschen series respectively then $\lambda_1 : \lambda_2$ is

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Answer: (d)

Lyman first line: $\frac{1}{\lambda_1} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}$ , so $\lambda_1 = \frac{4}{3R}$ .
Paschen first line: $\frac{1}{\lambda_2} = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$ , so $\lambda_2 = \frac{144}{7R}$ .
$\frac{\lambda_1}{\lambda_2} = \frac{4}{3} \times \frac{7}{144} = \frac{7}{108}$ .
So the answer is (d) $7 : 108$ .

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